Charging a capacitor through a high resistance

Please help me with this question. Thanks...
i read from book that in order to charge an uncharged capacitor, we have to connect the capacitor and a resistor with high resistance in series to a battery with emf E. I don't understand what is the use and the effect of the resistor in this charging process? I know that the time constant is given by the product of capacitance and resistance. But is it possible to charge the capacitor without the resistor?

Well.. charging and discharging both require a current to flow. And current will flow between two points only and only if there is a potential difference between the two points. The lead wires that we use for connections have a very low resistance.. further, these wires are assumed to have zero resistance by most of the textbooks. As such, current will not flow since there is no or a very small potential difference between the two points. To increase this difference, a resistor is added. The resistor causes a huge potential gradient in the circuit as compared to the lead wires and causes current to flow.

Another way to see this is that the amount of Energy dissipated [by Joules heating] is proportional to the square of the resistance. A charged capacitor holds some energy with it. In order to discharge, the energy must somehow be lost to the surroundings. To achieve this, we need some resistance in the circuit which will dissipate this energy in the form of heat. Higher the resistance, more is the power i.e. more energy is dissipated per unit time and hence a Capacitor will be discharged quickly.

Yes.. it is possible to charge the capacitor without a resistor. This is achieved by using a battery to create a potential difference.

Yes.. it is also possible to discharge the capacitor without a resistor by using a battery.. but it would cause the capacitor to eventually charge again.. but this time with a different polarity. For ex., if a capacitor has +q on it's right plate and -q on it's left plate, and we connect a battery [positive to left plate], then the charge of the capacitor will first go down to zero and then will again go back to q, but this time we will have -q on it's right plate and +q on it's left plate. However, it is very difficult to determine the exact moment when this charge becomes zero and disconnect it and hence we use a resistor for that purpose. This behavior may differ with rechargeable batteries [i'm not sure about that though].

Thank you very much. But i am confused that a charged capacitor will take a longer time or shorter time to discharge if the resistance in the circuit is bigger? "A larger resistance results in a smaller discharge current", is this correct?

Haha, that makes me think of a recent compulsory "electrical safety" course I had to follow (a few weeks ago). It's a silly thing, but in our place we are obliged (no matter what degree you hold) to follow every 3 years a compulsory course on safety, and one of the courses is on electricity ("don't put your fingers in the outlet" course, we call it).

And there was an initial multiple choice quiz handed out to have an idea of the class participants, and the first question was: does a thicker wire has a (a) higher (b) lower resistance than a thin wire ?

And lo and behold, we had a computer scientist there who answered (a). I tried to save his face by saying that maybe the question was about the *mechanical* resistance to breaking (think of Tarzan), which gained me the angry looks of the teacher for the rest of the morning session :rofl: